Question

(M) Use the method in this section to produce a $QR$ factorization of the matrix in Exercise 24.

Short answer

The required factorization is, $\left(\begin{array}{rlrl}r-10& 13& 7& -11\\ 2& 1& -5& 3\\ -6& 3& 13& -3\\ 16& -16& -2& 5\\ 2& 1& -5& -7\end{array}\right)=\left(\begin{array}{rlrl}r\frac{1}{2}& \frac{1}{2}& \frac{1}{\sqrt{3}}& 0\\ \frac{1}{10}& \frac{1}{2}& 0& \frac{1}{\sqrt{2}}\\ \frac{-3}{10}& -\frac{1}{2}& \frac{1}{\sqrt{3}}& 0\\ \frac{4}{5}& 0& \frac{1}{\sqrt{3}}& 0\\ \frac{1}{10}& \frac{1}{2}& 0& \frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{rlrl}20& -20& -10& 10\\ 0& 6& -8& -6\\ 0& 0& 6\sqrt{3}& -3\sqrt{3}\\ 0& 0& 0& 5\sqrt{2}\end{array}\right)$.

Step-by-step solution

$QR$ factorization of a Matrix

A matrix with order $m\times n$ can be written as the multiplication of an upper triangular matrix $R$ and a matrix $Q$ which is formed by applying the Gram–Schmidt orthogonalization processto the $\mathrm{c}\mathrm{o}\mathrm{l}\left(A\right)$.

The matrix $R$ can be found by the formula $Q^{T}A=R$.

Finding the matrix $R$

Given that, $A=\left(\begin{array}{rlrl}r-10& 13& 7& -11\\ 2& 1& -5& 3\\ -6& 3& 13& -3\\ 16& -16& -2& 5\\ 2& 1& -5& -7\end{array}\right)$.

Hence, enter matrix A in MATLAB.

>> A=(-10 13 7 -11; 2 1 5 3; -6 3 13 -3; 16 -16 -2 5; 2 1 -5 -7);

The required function:

function (B) = GramSchmidt_N(A)

(m,n) = size(A);

(U, jb) = rref(A);

x = length(jb);

B = zeros(m,x);

for i = 1:x

C(:,i)= A(:,(jb(i)));

end

B=C;

for i = 2:x

for j = 1:i-1

B(:,i) = B(:,i) - dot(C(:,i),B(:,j))/dot(B(:,j),B(:,j))* B(:,j)

end

end

for i=1:size(B,2)

TMP=B(:,i);

TMP=TMP./(sqrt(sum(TMP.^2)));

B(:,i)=TMP;

end

end

Find the Normalised orthogonal basis:

(B) = GramSchmidt_N(A)

$\begin{array}{r}B=\\ \begin{array}{rlrl}r-0.5000& 0.5000& 0.5774& 0\\ 0.1000& 0.5000& 0& 0.7071\\ -0.3000& -0.5000& 0.5774& 0\\ 0.8000& 0& 0.5774& 0\\ 0.1000& 0.5000& 0& -0.7071\end{array}\end{array}$

Therefore, the orthogonal matrix $Q$ is$Q=\left(\begin{array}{rlrl}r\frac{1}{2}& \frac{1}{2}& \frac{1}{\sqrt{3}}& 0\\ \frac{1}{10}& \frac{1}{2}& 0& \frac{1}{\sqrt{2}}\\ \frac{-3}{10}& -\frac{1}{2}& \frac{1}{\sqrt{3}}& 0\\ \frac{4}{5}& 0& \frac{1}{\sqrt{3}}& 0\\ \frac{1}{10}& \frac{1}{2}& 0& \frac{1}{\sqrt{2}}\end{array}\right)$.

Find the matrix $R$ using the following formula:

$R=Q^{T}A$

R=Q'*A

$\begin{array}{r}R=\\ \begin{array}{rlrl}20.0000& -20.0000& -10.0000& 10.0000\\ 0& 6.00000& -8.0000& -6.0000\\ 0& 0& 10.3923& -5.1962\\ 0& 0& 0& 7.0711\end{array}\end{array}$

Hence, the matrix is$R=\left(\begin{array}{rlrl}20& -20& -10& 10\\ 0& 6& -8& -6\\ 0& 0& 6\sqrt{3}& -3\sqrt{3}\\ 0& 0& 0& 5\sqrt{2}\end{array}\right)$.